2005 AMC 12B Problem 11

Question

An envelope contains eight bills:

2

ones,

2

fives,

2

tens, and

2

twenties. Two bills are drawn at random without replacement. What is the probability that their sum is

\$20

or more?

\dfrac{1}{4}

\dfrac{2}{7}

\dfrac{3}{7}

\dfrac{1}{2}

\dfrac{2}{3}

There are

\binom{8}{2} = 28

equally likely pairs

A sum of

\$20

or more needs both twenties, a twenty with a smaller bill, or both tens

There are

\binom{8}{2} = 28

equally likely pairs of bills.

The sum is

\$20

or more in these cases: both twenties (

1

way), one twenty with one of the six smaller bills (

2 \cdot 6 = 12

ways), or both tens (

1

way).

That is

1 + 12 + 1 = 14

favorable pairs, so the probability is

\dfrac{14}{28} = \dfrac12.

Thus, the correct answer is D.

Problem 11 in other years